Grade 10 math word problems with answers and solutions are presented.
Problems
A real estate agent received a 6% commission on the selling price of a house. If his commission was $8,880, what was the selling price of the house?
An electric motor makes 3,000 revolutions per minutes. How many degrees does it rotate in one second?
The area of a rectangular field is equal to 300 square meters. Its perimeter is equal to 70 meters. Find the length and width of this rectangle.
The area of the
trapezoid shown below is equal to 270 square units. Find its perimeter and round your answer to the nearest unit.
.
If a tire rotates at 400 revolutions per minute when the car is traveling 72km/h, what is the circumference of the tire?
In a shop, the cost of 4 shirts, 4 pairs of trousers and 2 hats is $560. The cost of 9 shirts, 9 pairs of trousers and 6 hats is $1,290. What is the total cost of 1 shirt, 1 pair of trousers and 1 hat?
Four children have small toys. The first child has 1/10 of the toys, the second child has 12 more toys than the first, the third child has one more toy of what the first child has and the fourth child has double the third child.
How many toys are there?
A class average mark in an exam is 70. The average of students who scored below 60 is 50. The average of students who scored 60 or more is 75. If the total number of students in this class is 20, how many students scored below 60?
For what value of x will the function f(x) = -3(x - 10)(x - 4) have a maximum value? Find the maximum value.
It takes a boat 3 hours to travel down a river from point A to point
B, and 5 hours to travel up the river from B to A. How long would it
take the same boat to go from A to B in still water?
An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.
Solutions to the Above Problems
6% x = 8,880 : x = selling price of house.
x = $148,000 : solve for x.
L × W = 300 : area , L is the length and W is the width.
2 L + 2 W = 70 : perimeter
L = 35 - w : solve for L
(35 - W) × W = 300 : substitute in the area equation
W = 15 and L = 20 : solve for W and find L using L = 35 - w.
Let h be the height of the trapezoid.
area = (1/2) × h × (10 + 10 + 3 + 4) = 270
h = 20 : solve for h
20^{2} + 3^{2} = L^{2} : Pythagora's theorem applied to the right triangle on the left.
L = sqrt(409)
20^{2} + 4^{2} = R^{2} : Pythagora's theorem applied to the right triangle on the right.
R = sqrt(416)
perimeter = sqrt(409) + 10 + sqrt(416) + 17 = 27 + sqrt(409) + sqrt(416)
400 rev / minute = 400 × 60 rev / 60 minutes
= 24,000 rev / hour
24,000 × C = 72,000 m : C is the circumference
C = 3 meters
Let x be the price of one shirt, y be the price of one pair of trousers and z be the price of one hat.
4x + 4y + 2z = 560 :
9x + 9y + 6z = 1,290
3x + 3y + 2z = 430 : divide all terms of equation C by 3
x + y = 130 : subtract equation D from equation B
3(x + y) + 2z = 430 : equation D with factored terms.
3*130 + 2z = 430
z = 20 : solve for z
x + y + z = 130 + 20 = $150
x : the total number of toys
x/10 : the number of toys for first child
x/10 + 12 : the number of toys for second child
x/10 + 1 : the number of toys for the third child
2(x/10 + 1) : the number of toys for the fourth child
x/10 + x/10 + 12 + x/10 + 1 + 2(x/10 + 1) = x
x = 30 toys : solve for x
Let n the number of students who scored below 60 and N the number of students who scored 60 or more. Xi the grades below 60 and Yi the grades 60 or above.
[sum(Xi) + sum(Yi)] / 20 = 70 : class average
sum(Xi) / n = 50 : average for less that 60
sum(Yi) / N = 75 : average for 60 or more
50n + 75N = 1400 : combine the above equations
n + N = 20 : total number of students
n = 4 and N = 16 : solve the above system
f(x) = -3(x - 10)(x - 4) = -3 x^{2} + 42 x - 120 : expand and obtain a quadratic function
h = -b/2a = - 42/(-6) = 7 : h is the value of x for which f has a maximum value
f(h) = f(7) = 27 : maximum value of f.
Let: S be the speed of the boat in still water, r be the rate of the water current and d the distance between A and B.
d = 3(S + r) : boat travelling down river
d = 5(S - r) : boat travelling up river
3(S + r) = 5(S - r)
r = S / 4 : solve above equation for r
d = 3(S + S/4) : substitute r by S/4 in equation B
d / S = 3.75 hours = 3 hours and 45 minutes.
Let: S be the speed of the airplane in still air, r be the speed of the wind and d the distance between A and B.
d = 8(S - r) : airplane flies against the wind
d = 7(S + r) : airplane flies in the same direction as the wind
8(S - r) = 7(S + r)
S/r = 15